If the current was less, you could simply power it from a PIC pin... I've done that before. They will source or sink up to 20mA.
A PNP transistor will give you a pretty good 'high side' switch, particularly for such a low current.
VCC + | 1k | ___ |< PIC -|___|--| PNP like 2N4403 or 2N3906 |\ | | Output '-------------
Drop is about 0.2V, which is Vce(sat) for the transistor.
(created by AACircuit v1.28 beta 10/06/04
Turn it *off* by setting the pin high. Turn it on by setting the pin to ground.
You can use a MOSFET as well. In that case, you don't need a gate resistor, and it's possible the V drop won't be as severe. However, make sure you get a 'logic level' MOSFET. Otherwise, you won't be able to turn it all the way on, and the drop across it will be bigger. Logic level mosfets turn on at a Vgs of less than 5V, as specified in the datasheet.
An N-channel mosfet will go between ground and your circuit, a P-channel mosfet will go between VCC and your circuit. N or NPN are both active high, whereas P and PNP are active low.
Reed switches may also work, but I don't have any experience with them, so I can't help there.